分配bash函数参数新值

Question

Is it possible to first pass a parameter to the function and than change its value?

#!/bin/bash

name=old_name

echo $name #echoes "old_name"

alter () {
$1=new_name #throws error that says 'command not found'
}

alter name

echo $name #I would like to see "new_name" here

Answer 1

Yes, using a nameref:

alter () {
    declare -n foo=$1
    foo=new_name
}

See Bash FAQ 006 for more advice and warnings, as well as workarounds for versions of bash that predate nameref support (ie, 4.2 or earlier).

Answer 2

You can't do it that way, unfortunately. Also, bash functions can't return values. The usual options are (1) set a global var in the function (yuck), or (2) echo the "return" value and use command substitution to call it. Something like this:

alter () {
  echo "$1 but different"
}

name="Fred"
name=$(alter $name)
echo Name is now $name

returns: Name is now Fred but different